$f(x, y) = x^2 - y^2 + x - y$ What is the partial derivative of $f$ with respect to $y$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-2y - 1$ (Choice B) B $2x + 1$ (Choice C) C $0$ (Choice D) D $2x - 2y$
Answer: Taking a partial derivative with respect to $y$ means treating $x$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial y} &= \dfrac{\partial}{\partial y} \left[ x^2 - {y^2} + x - {y} \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ x^2 \right] - \dfrac{\partial}{\partial y} \left[ {y^2} \right] + \dfrac{\partial}{\partial y} \left[ x \right] - \dfrac{\partial}{\partial y} \left[ {y} \right] \\ \\ &= 0 - {2y} + 0 - 1 \end{aligned}$ In conclusion, $\dfrac{\partial f}{\partial y} = -2y - 1$